1.3.2 Example 2. \(xy^{\prime }+y=x\)

\begin{equation} xy^{\prime }+y=x \tag {1}\end{equation}

This is same as example 1, but with nonzero on RHS. The solution is \(y=y_{h}+y_{p}\). Where \(y_{h}\) was found above as

\[ y_{h}=\frac {a_{0}}{x}\]

To find \(y_{p}\) we will use the balance equation, EQ (*) found in the first example when finding \(y_{h}\). We just need to rename \(a_{0}\) to \(c_{0}\) and add the \(x\) on the right side of the balance equation.

\[ \left ( r+1\right ) c_{0}x^{r}=x \]

For balance we see that

\[ r=1 \]

Hence

\begin{align*} \left ( r+1\right ) c_{0} & =1\\ 2c_{0} & =1\\ c_{0} & =\frac {1}{2}\end{align*}

Now that we found \(r,c_{0}\) we will use the summation equation in first example to find all \(c_{n}\) for \(n>0\,\). We see that all summations terms start from the same index. This implies that only term exist for \(y_{p}\) which is

\begin{align*} y_{p} & =c_{0}x^{r}\\ & =\frac {1}{2}x \end{align*}

If the summation equation did not have all the sums in it start from same lower index, then we would had to apply the summation equation to find all \(c_{n}\) for \(n>0\) just like we did for finding \(a_{n}\). But here we got lucky. Therefore

\begin{align*} y & =y_{h}+y_{p}\\ & =\frac {a_{0}}{x}+\frac {1}{2}x \end{align*}

Last problem below gives case when not all sums in the summation equation have the same starting index.